Circular motion — centripetal acceleration derivation: Worked examples

1.2.15 · D3 · Physics › Newton's Laws & Dynamics › Circular motion — centripetal acceleration derivation

Yeh page parent topic ka "kuch bhi poocho" companion hai. Physics settle ho chuki hai: koi object speed $v$ se radius $r$ ke circle par turn kare toh uska inward acceleration hota hai $a_{c} = \frac{v ^{2}}{r} = ω^{2} r = v ω .$ Yahan $v$ hai speed (metres per second, hamesha positive), $r$ hai radius (metres, hamesha positive), aur $ω$ (Greek letter "omega") hai angular speed — object ek second mein kitne radians ka angle sweep karta hai. Agar in teeno mein se koi word shaky lage, pehle Angular velocity and period padho, phir wapas aao.

Ek aur input word jo humein chahiye: frequency $f$ object kitne full turns complete karta hai har second, measured in hertz ( $Hz$ , yaani turns per second). Kyunki ek turn $2 π$ radians hota hai, angular speed aur frequency ka relation hai $ω = 2 π f$ . Frequency aur period reciprocals hain: $f = 1/ T$ .

Ab hamara kaam: har tarah ka problem enumerate karo jo yeh ek formula dress up ho sakta hai, phir har ek ka ek example karo taaki koi bhi scenario aisa na mile jise tumne pehle dekha nahi.

The scenario matrix

Har centripetal-acceleration problem basically yeh hai — "tumhe teen quantities mein se do di gayi hain aur teesri poochhi gayi hai". Neeche ki rows class of input list karti hain jo tumhe milta hai; last teen rows woh traps hain jo exams upar se add karte hain.

Case	Case class	Kya diya gaya hai	Trick jo tumhe pata honi chahiye
A	Speed + radius	$v, r$	seedha $v^{2} / r$ mein plug karo
B	Angular speed + radius	$ω, r$	$ω^{2} r$ use karo, pehle $v$ mat nikalo
C	Period or frequency	$T$ or $f$ , and $r$	$ω = 2 π / T = 2 π f$ mein convert karo
D	Ulta solve karo	$a_{c}$ pata hai, $v$ ya $r$ nikalo	formula rearrange karo
E	Degenerate: zero input	$v = 0$ ya $r \to \infty$	acceleration $\to 0$ (straight line)
F	Limiting behaviour	$v$ ya $r$ scale karo	$a_{c} \propto v^{2}$ , $a_{c} \propto 1/ r$
G	Real-world word problem	mixed units (km/h, rpm)	plug karne se pehle units convert karo
H	Exam twist: force + Newton II	mass, string breaks	$F_{c} = m a_{c}$ se link karo aur Newton's Second Law

Neeche ke nau examples A se H cells ko order mein hit karte hain (Case C ko do milte hain: ek period se, ek frequency se). Har ek batata hai ki woh kaunsa cell cover karta hai.

Sabke peeche ek hi picture hai

Figure — Circular motion — centripetal acceleration derivation

Figure dekho. Object circle par baitha hai. Usse do arrows nikalte hain: coral arrow velocity $v$ hai, hamesha tangent (circle ko graze karta hua, jis taraf ja raha hai usi taraf point karta). Lavender arrow acceleration $a_{c}$ hai, hamesha seedha centre ki taraf point karta, $v$ ke perpendicular. Har example bas yeh poochh raha hai — "lavender arrow kitna lamba hai?"

Case A — speed aur radius diye gaye hain

Case B — angular speed aur radius diye gaye hain

Worked example Whirling stone

Ek patthar $ω = 10 rad/s$ par $r = 0.5 m$ length ki string par ghoom raha hai. $a_{c}$ nikalo.

Forecast: Kya $ω^{2} r$ use karne par upar ki car se bada ya chhota number milega?

$a_{c} = ω^{2} r$ choose karo. Yeh step kyun? Humein directly angular speed $ω$ diya gaya hai. Pehle $v$ mein convert karna sirf extra kaam hoga (aur galti karne ka mauka).
$ω$ square karo: $ω^{2} = 1 0^{2} = 100 rad^{2} / s^{2}$ . Yeh step kyun? Same "squared" structure — do guna fast spin karne par pull chaar guna ho jata hai.
Radius se multiply karo: $a_{c} = 100 \times 0.5 = 50 m/s^{2}$ . Yeh step kyun? Yahan bada radius $a_{c}$ ko bada karta hai (kyunki $v = ω r$ bhi badhta hai) — Case A ke ulta, toh blindly mat maano "bada $r$ = chhota $a_{c}$ ".

Verify: $v$ se cross-check karo: $v = ω r = 10 \times 0.5 = 5 m/s$ , phir $v^{2} / r = 25/0.5 = 50 m/s^{2}$ ✓. Dono raaste mile. Units: radians dimensionless hain, toh $s^{- 2} \cdot m = m/s^{2}$ ✓.

Case C — period ya frequency di gayi hai

Worked example C1 — Satellite apne period se

Ek satellite $r = 7.0 \times 1 0^{6} m$ radius par period $T = 5800 s$ ke saath orbit karta hai. $a_{c}$ nikalo.

Forecast: Kya $a_{c}$ $9$ ke aas-paas hoga, ya $0.001$ jaisa chhota?

Period ko angular speed mein convert karo: $ω = \frac{2 π}{T}$ . Yeh step kyun? Ek full loop $2 π$ radians hota hai; har $T$ seconds mein ek baar karna matlab $2 π / T$ radians har second. Dekho Angular velocity and period.
Compute karo $ω = \frac{2 π}{5800} \approx 1.0834 \times 1 0^{- 3} rad/s$ . Yeh step kyun? Slow orbit ⇒ chhota $ω$ , jaisa expected hai kisi cheez ke liye jo ~1.6 hours per loop leta hai.
Apply karo $a_{c} = ω^{2} r = (1.0834 \times 1 0^{- 3})^{2} \times 7.0 \times 1 0^{6} \approx 8.22 m/s^{2}$ . Yeh step kyun? Hum period ko seedha chain kar sakte hain: $a_{c} = \frac{4 π ^{2} r}{T ^{2}}$ , jo bas $ω^{2} r$ hai jisme $ω = 2 π / T$ substitute kiya gaya hai.

Verify: $\frac{4 π ^{2} r}{T ^{2}} = \frac{4 π ^{2} ( 7.0 \times 1 0 ^{6} )}{580 0 ^{2}} \approx 8.22 m/s^{2}$ ✓. Sanity: yeh $g = 9.8$ ke close hai — sahi hai, low-orbit gravity almost poori strength par hoti hai (dekho Gravitation — orbital motion).

Worked example C2 — Fan blade tip apni frequency se

Ek ceiling fan frequency $f = 4 Hz$ (char turns per second) par spin karta hai. Blade par ek point axis se $r = 0.60 m$ door hai. $a_{c}$ nikalo.

Forecast: Satellite ke $8 m/s^{2}$ se bada ya chhota?

Frequency ko angular speed mein convert karo: $ω = 2 π f$ . Yeh step kyun? Har turn $2 π$ radians hota hai, aur $f$ turns per second hain, toh radians swept per second $2 π f$ hai. (Equivalently $ω = 2 π / T$ with $T = 1/ f$ .)
Compute karo $ω = 2 π \times 4 = 8 π \approx 25.13 rad/s$ . Yeh step kyun? Fast, small spin ⇒ bada $ω$ , slow satellite ke ulta.
Apply karo $a_{c} = ω^{2} r = (8 π)^{2} \times 0.60 = 4 π^{2} f^{2} r \approx 379 m/s^{2}$ . Yeh step kyun? $ω = 2 π f$ ko $ω^{2} r$ mein substitute karne par frequency form milta hai $a_{c} = 4 π^{2} f^{2} r$ — $f$ seedha plug karo.

Verify: $4 π^{2} f^{2} r = 4 π^{2} (4)^{2} (0.60) \approx 379 m/s^{2}$ ✓. $8$ se kaafi bada — fast spin thode radius ke paas bada win karta hai, jaisa forecast tha.

Case D — missing quantity ke liye ulta solve karo

Case E — degenerate inputs (zero aur infinity)

Worked example Do "boring" limits

(a) Ek merry-go-round ruka hua hai: $v = 0$ , $r = 3 m$ . $a_{c}$ nikalo. (b) Ek car bilkul seedhi road par chalti hai — ise $r \to \infty$ radius ke circle ki tarah treat karo $v = 30 m/s$ par. $a_{c}$ nikalo.

Forecast: Dono ko same physically-obvious number dena chahiye. Kaunsa?

Case (a): $a_{c} = v^{2} / r = 0^{2} /3 = 0 m/s^{2}$ . Yeh step kyun? Move nahi kar raha ⇒ velocity vector turn nahi kar raha ⇒ accelerate karne ko kuch nahi. Formula bhi agree karta hai.
Case (b): $a_{c} = v^{2} / r = 3 0^{2} / r$ ; $r$ ko bina bound ke badhne do. Yeh step kyun? Straight line infinite radius ka circle hai. Jaise $r \to \infty$ , koi bhi fixed numerator ek bade denominator par $\to 0$ .
Toh $r \to \infty lim \frac{900}{r} = 0 m/s^{2}$ . Yeh step kyun? Yeh exactly Newton's First Law (Inertia) hai: koi turning force nahi ⇒ straight line ⇒ zero centripetal acceleration.

Verify: Dono $0$ dete hain. Consistent: centripetal acceleration tab hi hota hai jab path actually curve kare. Ruka hua object aur infinitely-gentle curve dono curve karne mein fail karte hain, toh dono $0$ dete hain ✓.

Case F — limiting / scaling behaviour

Worked example Speed double karo, radius half karo

Ek cyclist corner par $a_{c} = 4 m/s^{2}$ se ghoomti hai. Doosri run mein woh apni speed double karti hai aur half radius ke corner par jaati hai. Naya $a_{c}$ kya hoga?

Forecast: ×2? ×4? ×8?

Numbers nahi, ratio likho: $\frac{a _{c, new}}{a _{c, old}} = \frac{v _{new}^{2} / r _{new}}{v _{old}^{2} / r _{old}}$ . Yeh step kyun? Humein actual $v$ ya $r$ kabhi nahi bataya gaya, sirf kaise change hote hain — toh ratios se kaam karo aur sab unknown cancel ho jaate hain.
Speed double ⇒ $v_{new} = 2 v_{old}$ ⇒ $v^{2}$ factor $2^{2} = 4$ hai. Radius half ⇒ $r_{new} = \frac{1}{2} r_{old}$ ⇒ half se divide karna matlab ×2. Yeh step kyun? $a_{c} \propto v^{2}$ aur $a_{c} \propto 1/ r$ independently kaam karte hain; unke factors multiply karo.
Total factor $= 4 \times 2 = 8$ , toh $a_{c, new} = 8 \times 4 = 32 m/s^{2}$ . Yeh step kyun? Dono changes turn ko harder banate hain, toh pile up ho jaate hain — isliye "thodi si tez speed tighter bend par" cars ko spin out kar deti hai.

Verify: Ek concrete case lo: $v_{old} = 4 m/s, r_{old} = 4 m \Rightarrow a = 16/4 = 4$ ✓. Naya $v = 8, r = 2 \Rightarrow 64/2 = 32 m/s^{2}$ ✓. Ratio $32/4 = 8$ ✓.

Case G — real-world word problem (messy units)

Worked example Turntable rpm mein, kitchen-table radius

Ek vinyl record $33 \frac{1}{3} rpm$ (revolutions per minute) par spin karta hai. Dust ka ek tukda centre se $r = 0.15 m$ door baitha hai. Uska $a_{c}$ nikalo.

Forecast: $2 m/s^{2}$ se bada ya chhota?

rpm ko rad/s mein convert karo: $ω = 33.333 \frac{rev}{min} \times \frac{2 π rad}{1 rev} \times \frac{1 min}{60 s}$ . Yeh step kyun? Formula sirf radians-per-second samajhta hai. Pehle convert karo, phir plug karo — rpm ko kabhi seedha mat daalo.
Compute karo: $ω = \frac{33.333 \times 2 π}{60} \approx 3.491 rad/s$ . Yeh step kyun? Ek revolution $2 π$ radians hota hai; sixty seconds per minute "per minute" ko "per second" mein convert karta hai.
Apply karo $a_{c} = ω^{2} r = (3.491)^{2} \times 0.15 \approx 1.828 m/s^{2}$ . Yeh step kyun? Humein spinning ki rate (rpm ⇒ $ω$ ) aur radius diya gaya hai, toh Case-B form $ω^{2} r$ natural hai.

Verify: Alternative route: $v = ω r = 3.491 \times 0.15 = 0.5237 m/s$ , phir $v^{2} / r = 0.523 7^{2} /0.15 \approx 1.828 m/s^{2}$ ✓. $2 m/s^{2}$ se neeche jaisa forecast tha — gentle, dust tika rehta hai.

Case H — exam twist: force, Newton II, aur ek breaking string

Algebra se pehle figure padho. Butter-yellow line string hai; coral dot ball hai. Sirf ek arrow andar ki taraf point karta hai — lavender tension arrow $T$ — aur wahi ek inward force hi centripetal force $F_{c}$ hai (koi alag "centripetal force" add karne ki zaroorat nahi). Mint arrow ball ki velocity $v$ hai, dashed circle ke tangent. Poora example yeh hai: tension sirf itni badi ho sakti hai jab tak string snap nahi ho jaata, aur wahi cap karta hai kitni fast hum whirl kar sakte hain.

Worked example String kab tak snap nahi hogi?

$m = 0.20 kg$ ka ek ball horizontal string of length $r = 0.80 m$ par whirl kiya jata hai. String tab toot jaati hai jab tension $F_{m a x} = 40 N$ se zyada ho. Maximum speed $v_{m a x}$ kya hai?

Forecast: $8$ , $12$ , ya $40 m/s$ ke aas-paas?

Identify karo ki centripetal force kya provide kar raha hai. Yahan tension hi sirf inward force hai, toh $F_{c} = T$ . Yeh step kyun? Centripetal force kabhi naya force nahi hota — yeh real forces se supplied net inward force hai (parent note ki mistake list dekho). Is horizontal whirl mein, woh string tension hai (figure mein lavender arrow).
Radial direction mein Newton's Second Law apply karo: $T = m a_{c} = \frac{m v ^{2}}{r}$ . Yeh step kyun? Newton II kehta hai net force $=$ mass × acceleration; inward net force tension ke barabar hai, acceleration $v^{2} / r$ hai.
String tab tooti hai jab $T = F_{m a x}$ , toh set karo $\frac{m v _{m a x}^{2}}{r} = F_{m a x}$ aur solve karo: $v_{m a x} = \frac{F _{m a x} r}{m}$ . Yeh step kyun? Unknown speed ke liye rearrange karo; string jo maximum tension hold kar sakti hai woh maximum speed set karta hai.
Numbers: $v_{m a x} = \frac{40 \times 0.80}{0.20} = 160 \approx 12.65 m/s$ . Yeh step kyun? Plug in karo; square root $v^{2}$ ko undo karta hai.

Verify: Back-substitute karo: $a_{c} = v_{m a x}^{2} / r = 160/0.80 = 200 m/s^{2}$ , phir $F = m a_{c} = 0.20 \times 200 = 40 N = F_{m a x}$ ✓. $v$ ki units: $\frac{N \cdot m}{kg} = \frac{kg m/s ^{2} \cdot m}{kg} = m^{2} / s^{2} = m/s$ ✓.

Wrap-up recall

Recall Kaunse input ke liye kaunsa formula?

$v$ aur $r$ diye gaye hain ::: $a_{c} = v^{2} / r$ use karo. $ω$ aur $r$ diye gaye hain ::: $a_{c} = ω^{2} r$ use karo (pehle $v$ mein convert mat karo). Period $T$ diya gaya hai ::: $a_{c} = 4 π^{2} r / T^{2}$ use karo (kyunki $ω = 2 π / T$ ). Frequency $f$ di gayi hai ::: $a_{c} = 4 π^{2} f^{2} r$ use karo (kyunki $ω = 2 π f$ ). Breaking-string tension di gayi hai ::: us tension ko $m v^{2} / r$ ke barabar set karo Newton II se. Object rest par ya straight road par ::: $a_{c} = 0$ (koi curving path nahi).